f(x) = x^3 - 2x^2 - 4x + 8 / x - 2
2006-09-29 07:09:41 · 8 answers · asked by pnoiz1 2 in Science & Mathematics ➔ Mathematics
(x^3-2x^2-4x+8)/(x-2)
=[x^2(x-2)-4(x-2)]/(x-2)
=(x^2-4)(x-2)/(x-2)
=(x^2-4)
=(x-2)(x+2)
2006-09-29 07:17:58 · answer #1 · answered by ag_iitkgp 7 · 0⤊ 0⤋
Fx=x^3-2x^2-(4x+8)/(x-2)
=x^2(x-2)-4(x-2)/(x-2)
=x^2(x-2)-4
2006-09-29 14:16:25 · answer #2 · answered by openpsychy 6 · 0⤊ 0⤋
f(x) = (x^3 - 2x^2 - 4x + 8)/(x - 2)
f(x) = ((x^3 - 2x^2) + (-4x + 8))/(x - 2)
f(x) = (x^2(x - 2) - 4(x - 2))/(x - 2)
f(x) = ((x^2 - 4)(x - 2))/(x - 2)
also you can factor this further to
f(x) = ((x - 2)(x + 2)(x - 2))/(x - 2)
only one of the (x - 2)s cancel out, which leaves you with
f(x) = x^2 - 4
2006-09-29 17:30:49 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
The answer should be (x^2 - 4) (x -2 )
2006-09-29 14:18:59 · answer #4 · answered by Tsuki 2 · 0⤊ 0⤋
is that 8/x that I see there? as in 8 DIVIDED by x? because if so, all the answers you got so far are incorrect - your expression is undefined in zero, while their is defined. Also, 2 is not a zero of this expression in this case.
2006-09-29 14:39:02 · answer #5 · answered by No Mo 2 · 0⤊ 0⤋
use horner method
for x=2 the numerator is zero so x=2 is a root.
the coefficients
........ 1 -2 -4 8
2......... 2 0 -8
----------------------------------------------------
1 0 -4 0
answer: f(x) = x^2-4
2006-09-29 14:32:52 · answer #6 · answered by iyiogrenci 6 · 0⤊ 0⤋
It is already in its simplest form and can't be simplified any further!
2006-09-29 14:12:33 · answer #7 · answered by quidwai 4 · 0⤊ 0⤋
is that a new way to do your homework??? don't cheat! put your own head at work!
2006-09-29 14:15:42 · answer #8 · answered by Pivoine 7 · 0⤊ 0⤋